Prove that $M/Tor(M) $ is torsion-free. Suppose $M$ is an $R$-module where $R$ is an integral domain.Define $Tor(M)$ be the set containing torsion elements of $M$. Prove that $M/Tor(M) $ is torsion-free. 
I have manage to prove that $Tor(M)$ is a submodule of $M$. Then my aim is to prove $Tor(M/Tor(M)) \cong \lbrace Tor(M) \rbrace$
My attempt: Let $m + Tor(M) \in Tor(M/Tor(M))$. Then there exists an $r \in R$, $r \neq 0$ such that $r(m+Tor(M))=rm+Tor(M)=Tor(M) \Rightarrow rm \in Tor(M)$. Then there exists an $s \in R,s \neq 0$ such that $srm=0$. Hence, we have $sr \neq 0 \Rightarrow m \in Tor(M)$. This tells us that all torsion elements of $M/Tor(M)$ is of the form $Tor(M)$, which means $Tor(M/Tor(M)) \subset \lbrace Tor(M) \rbrace$.
I don't know how to prove another direction. Can anyone help me?
 A: The identity element of a group and the zero element of a module are both torsion elements. In the second case one may multiply the zero element by any nonzero scalar of the ring (say $1$) and get zero, making zero a torsion element. The first case uses similar reasoning. This tells us that the zero element is always contained in the torsion subset of any module. In particular this allows us to go further and prove that the torsion subset is actually a submodule (for, how could we accept it is a submodule without believing it has the zero element?).
When we take any group, module or ring quotient $A/I$, the trivial coset $I$ itself is the identity or zero element of the quotient structure. In particular ${\rm Tor}(M)$ is the zero of $M/{\rm Tor}(M)$, so it is automatically contained in ${\rm Tor}(M/{\rm Tor}(M))$. Usually the facts that zero is torsion and that the torsion subset is actually a submodule are taken for granted and go without saying, so in a proof you would only need to show one direction: that assuming $x\in{\rm Tor}(M/{\rm Tor}(M))$ leads to the conclusion $x={\rm Tor}(M)$ (not "of the form," actually equal), which is what you've done already.
